Solutions to HW 10 Problems and Exercises: 37.28.Visualize:At t tt0 s, the origins of the S, S, and Sreference frames coincide. Solve:We have 1122221( / )1(0.80)1.667.v c(a) Using the Lorentz transformations, 8()1.6671200 m(0.80)(310m/s)(2.010s)xxvt61200 m 86228(0.80)(310m/s)(1200 m)1.667 2.010s2.0 s310m/svxttc (b)Using v c, the above equations yield x 2800 mand 8.67 s.t 37.29.Solve:We have 112222110.601.25.v cIn the earth’s reference frame, the We assume that the axes of the two frames coincide at t=t’= 0 sec Lorentz transformations yield 1081010810228()1.253.010m0.603.010m/s2008.2510m8.310m0.603.010m/s3.010m1.25200325 s330 s3.010m/sxxvtv xttc37.30.Model:S is the ground’s frame and Sis the rocket’s frame. Smoves with velocity v0.5crelative to S. Solve:(a)We have 1122221/10.501.155.v cApplying the Lorentz transformations to the lightning strike at x0 m and t10 s, 8552(1.155)0 m0.53.010m/s110s1732 m1700 m(1.155)(1 10s0 s)11.55 s12 sxxvtvxttc For the lightning strike at x30 km and t10 s, 485845281.1553.010m0.503.010m/s110s32.91 m33 m0.503.010m/s3.010m1.1551 10s46.2 s46 s3.010m/sxt (b)The events in the rocket’s frame are not simultaneous. The lightning is observed to strike the pole before the tree by 46 12 58 s. Hence we could infer, from direct application of Lorentz transformation (which is less confusing and more universal ), that simultaneity is relative.
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